# How can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=0$ using symmetry arguments?

Consider two integrals of the form $$I_{xx}=\int x^2 f(r)d^3r,~~I^\prime=\int xy ~f(r)d^3r$$ where $$f(r)$$ is a function of $$r=|\vec{r}|$$ only and has no dependence on $$\theta,\phi$$ in pherical polar coordinates. I have two questions. Can we show that $$I_{yy}=I_{zz}=I_{xx}$$ and $$I_{xy}=I_{yz}=I_{zx}=0$$ using symmetry arguments?

• Are you thrice integrating over $r$? Your question would be better answered if you clarify what exactly $I_{yy},I_{zz},I_{xy},I_{yz},I_{zx}$ are...... – Mostafa Ayaz Jan 6 at 12:33

Since $$f = f(|{\bf r}|)$$ it does not depend on the angles $$(\theta, \phi)$$, so at a given radius $$f$$ is a constant over the sphere defined by such radius. Now consider the term $$x$$, over the same sphere, for each $$x$$ there will be a $$-x$$, so for each $$x f$$ there will be a a $$-x f$$, the result after adding them up all together is zero. You can extend the argument to show that
$$I_{ij} = \int x_i x_j f(r) {\rm d}^3{\bf r} = 0 ~~~\mbox{for}~~i\not= j$$
And from this is also easy to see why $$I_{ii} \not = 0$$
$$f(r)$$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $$\int x^n f(r) d^3r=\int y^n f(r) d^3r=\int z^n f(r) d^3r$$. explicitly $$f(\sqrt{x^2+y^2+z^2})=f(\sqrt{y^2+x^2+z^2})$$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).
From the reflection symmetry we conclude that $$\int xy f(r) d^3r=\int (-x)y f(r) d^3r=-\int xy f(r) d^3r$$ so that $$I_{xy}=-I_{xy}=0$$, since $$f(\sqrt{x^2+y^2+z^2})=f(\sqrt{(-x)^2+y^2+z^2})$$ etc.